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JEE Mains · Physics · STD 12 -Electromagnetic induction
  • JEE Mains 2023
  • Hard
  • Numerical

A square loop of side \(2.0\,cm\) is placed inside a long solenoid that has \(50\) turns per centimetre and carries a sinusoidally varying current of amplitude \(2.5\,A\) and angular frequency \(700\,rad\,s ^{-1}\). The central axes of the loop and solenoid coincide. The amplitude of the emf induced in the loop is \(x \times 10^{-4} V\). The value of \(x\) is \(.........\)(Take, \(\pi=\frac{22}{7}\))

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Answer & Solution

Correct Answer

(C) \(44\)

Step-by-step Solution

Detailed explanation

\(B _{\text {due to solenoid }}=\mu_0 nI\)

\(\Phi_{\text {through square }}=\mu_0 nI \times A \quad( A =\text { Area })\)

\(Emf =\mu_0 nA \times \frac{ dI }{ dt }\)

\(=\mu_0 n A \times I _0 \omega \cos \omega t\)

\(\text { Emf amplitude }=\mu_0 n A \times I _0 \omega\)

\(=4 \pi \times 10^{-7} \times \frac{50}{10^{-2}} \times 4 \times 10^{-4} \times 2.5 \times 700\)

\(=44 \times 10^{-4}\,V\)

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